Let $F: \mathbb{R}^2 \rightarrow \mathbb{R}^4$ be defined by $F(u,v) = (u+v, uv, u-v, v^3)$ and let $M = F(\mathbb{R}^2)$. Prove that $M$ is a smooth manifold.
The proof that my TA posted online was this:
If the differential $DF|_{(u,v)}: T_{(u,v)} \mathbb{R}^2 \rightarrow T_{F(u,v)} \mathbb{R}^4$ is non degenerate then $F$ defines a smooth immersion of $\mathbb{R}^2$ into $\mathbb{R}^4$ and hence $F$ defines local diffeomorphisms for the image $M$. To verify that $F$ is a smooth immersion, note that $\mathrm{rank}(F) = 2 = \dim(\mathbb{R}^2)$. Because $F$ is globally injective, then it is actually a global diffeomorphism onto its image $M$. Hence $M$ is a smooth $2$-manifold.
I am struggling to understand this proof. I get why $F$ is a smooth immersion, but I don't get why it defines local diffeomorphisms. It seems as if he is applying the inverse function theorem to $F: \mathbb{R}^2 \rightarrow \mathbb{R}^4$, but he can't do this because the domain and co-domain do not have the same dimension. You could apply the inverse function theorem to $F: \mathbb{R}^2 \rightarrow M$, but this would require you to already know that $M$ is a smooth $2$-manifold, which is what we are trying to prove. What am I missing here? I would ask my TA but classes are done for the semester, so I won't be seeeing him again.
Best Answer
The proof your TA gave is incomplete: contrary to a widespread misconception a smooth injective immersion $F$ need not have as image a smooth manifold.
For two counterexamples look at Lee's Intoduction to Smooth Manifolds, Second Edition, Examples 4.19 (the notorious plane figure eight aka lemnicate) and Example 4.20, both on page 86.
The image of $F$ will however be a submanifold if $F$ is a proper injective immersion : same book, Proposition 4.22, page 87 (and preceding pages for excellent background).
Fortunately, in your case $F$ is indeed proper, as already demonstrated by its first and third components $u+v, u-v$ .